Question

Find the area of the region bounded by the graphs of the equations. Use a graphing utility to verify your result. $$ y=2 \sec \frac{\pi x}{6}, x=0, x=2, y=0 $$

Short answer

The area of the region bounded by the graphs of the given equations is \(\frac{12}{\pi}\ln(2+\sqrt{3})\) square units.

Step-by-step solution

Understand the functions

Recognize that the secant function \(y=2 \sec \frac{\pi x}{6}\) is just \(y=2/\cos(\pi x/6)\). Note that \(\cos(0)=1\) and \(\cos(\pi/3)=0.5\). Thus, the secant function starts at \(2\) when \(x=0\) and goes to \(4\) at \(x=2\). This function is bounded by the lines \(y=0\), \(x=0\) and \(x=2\).

Setup the integral for Area

The area \(A\) under the curve from \(x=a\) to \(x=b\) is given by \(A = \int_{a}^{b} f(x) dx\). Here, the area of region bounded by the graphs of the equations is given by \(A = \int_{0}^{2} 2 \sec \frac{\pi x}{6} dx\).

Evaluate the Integral

We can solve this integral in terms of \(\sec\), \(\tan\) or convert it in terms of \(\cos\), \(\sin\). Use a substitution, e.g. let \(u=\pi x/6\). Then, \(x=6u/\pi\) and \(dx=6 du/\pi\). Next, the integral becomes \(A= \frac{12}{\pi}\int_{0}^{\pi/3} \sec u du\). The integral of \(\sec u\) is \(\ln|\sec u + \tan u|\), to evaluate this integral between \(0\) and \(\pi/3\), we get the result \(\frac{12}{\pi}(\ln|\sec(\pi/3)+\tan(\pi/3)|-\ln|\sec(0)+\tan(0)|)\). Evaluating \(\sec\) and \(\tan\) at those points, we find that \(\sec(0)=1\), \(\tan(0)=0\), \(\sec(\pi/3)=2\), and \(\tan(\pi/3)=\sqrt{3}\). Substituting these in gives the final expression for the area: \(A=\frac{12}{\pi}(\ln|2+\sqrt{3}|-0) = \frac{12}{\pi}\ln(2+\sqrt{3})\) square units.

Verify Using Graphing Utility

At this step a graphing utility can be used to verify the area obtained. By plotting \(y=2 \sec \frac{\pi x}{6}\), \(x=0\), \(x=2\), and \(y=0\) one should expect to see a region with area appearing to be consistent with the value obtained through integration.